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Easy Java Simulations (2001- ) => kinematics => Topic started by: ahmedelshfie on June 01, 2010, 08:20:32 pm



Title: Cycloidal Pendulum
Post by: ahmedelshfie on June 01, 2010, 08:20:32 pm
This following applet is Cycloidal Pendulum
Created by prof Hwang Modified by Ahmed
Original project Cycloidal Pendulum (http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=1192.0)

A cycloid is the curve defined by the path of a point on the edge of circular wheel as the wheel rolls along a straight line. It is an example of a roulette, a curve generated by a curve rolling on another curve.

The cycloid is the solution to the brachistochrone problem (i.e. it is the curve of fastest descent under gravity) and the related tautochrone problem (i.e. the period of a ball rolling back and forth  inside this curve does not depend on the ball’s starting position).

The cycloid through the origin, generated by a circle of radius r, consists of the points (x, y), with

   $x = r(\omega t - \sin \omega t)$
   $y = r(1 - \cos \omega t)$
If its length is equal to that of half the cycloid, the bob of a pendulum suspended from the cusp of an inverted cycloid, such that the "string" is constrained between the adjacent arcs of the cycloid, also traces a cycloid path. Such a cycloidal pendulum is isochronous, regardless of amplitude. This is because the path of the pendulum bob traces out a cycloidal path (presuming the bob is suspended from a supple rope or chain); a cycloid is its own involute curve, and the cusp of an inverted cycloid forces the pendulum bob to move in a cycloidal path.

$v_x=\frac{dx}{dt}=r\omega*(1+\cos\omega t)$
$v_y=\frac{dy}{dt}=r\omega \sin \omega t$
Combined the above two equations: $(v_x-r\omega)^2+v_y^2)=(r\omega)^2$
So $v_x=\frac{v_x^2+v_y^2}{2r\omega}=\frac{2g\Delta y}{2r\omega}=\frac{g\Delta y}{r\omega}$
and $\frac{dy}{dx}=\frac{v_y}{v_x}=\frac{\sin\omega t}{1+\cos\omega t}$

The time required to travel from the top of the cycloid to the bottom is $T_{1/4}=\sqrt\frac{r}{g}\pi$

$a_x=\frac{d^2x}{dt^2}=-r\omega^2\sin\omega t$
$a_y=\frac{d^2y}{dt^2}= r\omega^2\cos\omega t$


Title: Re: Cycloidal Pendulum
Post by: ahmedelshfie on June 25, 2010, 08:20:06 pm
A cycloid is the curve defined by the path of a point on the edge of circular wheel as the wheel rolls along a straight line. It is an example of a roulette,
A curve generated by a curve rolling on another curve.
The cycloid is the solution to the brachistochrone problem (i.e. it is the curve of fastest descent under gravity) and the related tautochrone problem (i.e. the period of an object in descent without friction inside this curve does not depend on the ball's starting position).


Title: Re: Cycloidal Pendulum
Post by: ahmedelshfie on June 25, 2010, 08:21:50 pm
History

The cycloid was first studied by Nicholas of Cusa and later by Mersenne. It was named by Galileo in 1599. In 1634 G.P. de Roberval showed that the area under a cycloid is three times the area of its generating circle. In 1658 Christopher Wren showed that the length of a cycloid is four times the diameter of its generating circle. The cycloid has been called "The Helen of Geometers" as it caused frequent quarrels among 17th century mathematicians.
Data and images from http://en.wikipedia.org/wiki/Cycloid